Skip to main content

pattern matching - position of sequence of elements in list




Possible Duplicate:

Finding a subsequence in a list



Question


The position of {3, 5} is the list


{1, 3, 4, 3, 5, 5, 1}

is 3. How can such a position be found fast when dealing with large lists?


Attempt


This is what I could think of:


list = Table[RandomInteger[3], {200}]

ToString[FromDigits[%]]
StringPosition[%, "123"][[1, All]]

Out put example:


{1, 3, 2, 3, 1, 2, 2, 0, 0, 1, 0, 2, 3, 0, 2, 1, 2, 2, 3, 3, 1, 0, 1, 
 3, 1, 1, 0, 3, 1, 3, 2, 2, 2, 1, 2, 0, 2, 0, 2, 2, 3, 1, 3, 1, 1, 1, 
1, 3, 3, 1, 3, 1, 0, 2, 3, 1, 0, 3, 2, 3, 0, 1, 1, 3, 3, 3, 2, 1, 3, 
0, 1, 0, 1, 0, 3, 1, 1, 2, 0, 0, 2, 0, 1, 3, 1, 2, 0, 2, 0, 2, 2, 2, 
2, 2, 3, 1, 0, 3, 1, 2, 0, 3, 3, 2, 3, 1, 2, 3, 0, 0, 1, 2, 1, 2, 3, 
2, 0, 1, 3, 1, 1, 1, 3, 2, 3, 3, 2, 0, 2, 3, 0, 3, 0, 3, 3, 2, 3, 2, 

3, 1, 0, 1, 3, 0, 1, 1, 2, 2, 1, 2, 3, 3, 2, 3, 0, 2, 0, 3, 3, 2, 2, 
0, 2, 3, 3, 2, 2, 0, 2, 2, 2, 3, 1, 3, 2, 0, 2, 0, 3, 3, 1, 0, 1, 2, 
1, 2, 0, 0, 1, 0, 1, 0, 0, 2, 0, 3, 2, 1, 2, 2}

"132312200102302122331013110313222120202231311113313102310323011333213
0101031120020131202022222310312033231230012123201311132332023030332323
1013011221233230203322023322022231320203310121200101002032122"

{106, 108}


My solution does not work if the numbers in the list exceed 9 because there is no distinction between {..., 1, 2,...} and {...., 12, ....}. I also run into trouble if the sequence happens to start with 0.


I can fix this, but I doubt whether it is worth spending time on. There is likely a method that is faster in principle.



Answer



This isn't particularly fast, but I like it for simplicity:


seqpos[a_List, seq_List] := 
ReplaceList[a, {x___, Sequence @@ seq, ___} :> 1 + Length[{x}]]

seqpos[{1, 2, 3, 4, 2, 3}, {2, 3}]
(* {2,5} *)

Comments

Post a Comment

Popular posts from this blog

mathematical optimization - Minimizing using indices, error: Part::pkspec1: The expression cannot be used as a part specification

I want to use Minimize where the variables to minimize are indices pointing into an array. Here a MWE that hopefully shows what my problem is. vars = u@# & /@ Range[3]; cons = Flatten@ { Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; Minimize[{Total@((vec1[[#]] - vec2[[u[#]]])^2 & /@ Range[1, 3]), cons}, vars, Integers] The error I get: Part::pkspec1: The expression u[1] cannot be used as a part specification. >> Answer Ok, it seems that one can get around Mathematica trying to evaluate vec2[[u[1]]] too early by using the function Indexed[vec2,u[1]] . The working MWE would then look like the following: vars = u@# & /@ Range[3]; cons = Flatten@{ Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; NMinimize[ {Total@((vec1[[#]] - Indexed[vec2, u[#]])^2 & /@ R...
Post a Comment
Read more

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...
Post a Comment
Read more

What is and isn't a valid variable specification for Manipulate?

I have an expression whose terms have arguments (representing subscripts), like this: myExpr = A[0] + V[1,T] I would like to put it inside a Manipulate to see its value as I move around the parameters. (The goal is eventually to plot it wrt one of the variables inside.) However, Mathematica complains when I set V[1,T] as a manipulated variable: Manipulate[Evaluate[myExpr], {A[0], 0, 1}, {V[1, T], 0, 1}] (*Manipulate::vsform: Manipulate argument {V[1,T],0,1} does not have the correct form for a variable specification. >> *) As a workaround, if I get rid of the symbol T inside the argument, it works fine: Manipulate[ Evaluate[myExpr /. T -> 15], {A[0], 0, 1}, {V[1, 15], 0, 1}] Why this behavior? Can anyone point me to the documentation that says what counts as a valid variable? And is there a way to get Manpiulate to accept an expression with a symbolic argument as a variable? Investigations I've done so far: I tried using variableQ from this answer , but it says V[1...
Post a Comment
Read more